A. Kyoya and Photobooks
Kyoya Ootori is selling photobooks of the Ouran High School Host Club. He has 26 photos, labeled "a" to "z", and he has compiled them into a photo booklet with some photos in some order (possibly with some photos being duplicated). A photo booklet can be described as a string of lowercase letters, consisting of the photos in the booklet in order. He now wants to sell some "special edition" photobooks, each with one extra photo inserted anywhere in the book. He wants to make as many distinct photobooks as possible, so he can make more money. He asks Haruhi, how many distinct photobooks can he make by inserting one extra photo into the photobook he already has?
Please help Haruhi solve this problem.
The first line of input will be a single string s (1 ≤ |s| ≤ 20). String s consists only of lowercase English letters.
Output a single integer equal to the number of distinct photobooks Kyoya Ootori can make.
Note
In the first case, we can make 'ab','ac',...,'az','ba','ca',...,'za', and 'aa', producing a total of 51 distinct photo booklets.
#include <stdio.h>
#include <string.h> int main()
{
char str[];
int len;
scanf("%s",str);
len = strlen(str);
printf("%d\n",*(len+)-len); return ;
}
B. Ohana Cleans Up
Ohana Matsumae is trying to clean a room, which is divided up into an n by n grid of squares. Each square is initially either clean or dirty. Ohana can sweep her broom over columns of the grid. Her broom is very strange: if she sweeps over a clean square, it will become dirty, and if she sweeps over a dirty square, it will become clean. She wants to sweep some columns of the room to maximize the number of rows that are completely clean. It is not allowed to sweep over the part of the column, Ohana can only sweep the whole column.
Return the maximum number of rows that she can make completely clean.
The first line of input will be a single integer n (1 ≤ n ≤ 100).
The next n lines will describe the state of the room. The i-th line will contain a binary string with n characters denoting the state of the i-th row of the room. The j-th character on this line is '1' if the j-th square in the i-th row is clean, and '0' if it is dirty.
The output should be a single line containing an integer equal to a maximum possible number of rows that are completely clean.
Note
In the first sample, Ohana can sweep the 1st and 3rd columns. This will make the 1st and 4th row be completely clean.
In the second sample, everything is already clean, so Ohana doesn't need to do anything.
题意：每次改变一列的值（0变1，1变0）问最后最大有多少行全为1；
思路：不需要管全一或者全零因为他们是可以相互转换的，所以只要比较初始状态，每行状态，取最多的就好了；
题目链接：http://codeforces.com/contest/554/problem/B
转载请注明出处：寻找&星空の孩子
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std; char a[][]; int main()
{
int n,i,j,ans = ;
scanf("%d",&n);
for(i = ;i<n;i++)
{
scanf("%s",a[i]);
}
for(i = ;i<n;i++)
{
int s = ;
for(j = ;j<n;j++)
{
if(strcmp(a[i],a[j])==)
s++;
}
ans = max(ans,s);
}
printf("%d\n",ans);
}
C. Kyoya and Colored Balls
Kyoya Ootori has a bag with n colored balls that are colored with k different colors. The colors are labeled from 1 to k. Balls of the same color are indistinguishable. He draws balls from the bag one by one until the bag is empty. He noticed that he drew the last ball of color ibefore drawing the last ball of color i + 1 for all i from 1 to k - 1. Now he wonders how many different ways this can happen.
The first line of input will have one integer k (1 ≤ k ≤ 1000) the number of colors.
Then, k lines will follow. The i-th line will contain ci, the number of balls of the i-th color (1 ≤ ci ≤ 1000).
The total number of balls doesn't exceed 1000.
A single integer, the number of ways that Kyoya can draw the balls from the bag as described in the statement, modulo 1 000 000 007.
3
2
2
1
3
4
1
2
3
4
1680
In the first sample, we have 2 balls of color 1, 2 balls of color 2, and 1 ball of color 3. The three ways for Kyoya are:
1 2 1 2 3
1 1 2 2 3
2 1 1 2 3
题意：n种不同颜色的球，有k[n]个，要求每种颜色的球的最后一个的相对初始状态（球的种类）的位置不变；问有多少种组合
思路：定最后一个球，其他的球(相同的颜色)在前面任选,之后再定最后第二种颜色的球<放在剩下空中离最后一个位置最近的地方>，然后剩下的任选。。。以此类推。
题目链接：http://codeforces.com/contest/554/problem/C
转载请注明出处：寻找&星空の孩子
用了个费马小定理优化了下，也不可不优化。(a=1 mod (p-1),gcd(a,p)=1)
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
#define LL long long
const LL mod = ;
LL n;
LL a[];
LL fac[]; LL ppow(LL a,LL b)
{
LL c=;
while(b)
{
if(b&) c=c*a%mod;
b>>=;
a=a*a%mod;
}
return c;
} LL work(LL m,LL i)
{
return ((fac[m]%mod)*(ppow((fac[i]*fac[m-i])%mod,mod-)%mod))%mod;
} int main()
{
LL i,j,k;
fac[] = ;
for(i = ; i<; i++)
fac[i]=(fac[i-]*i)%mod;
LL ans = ,sum = ;
scanf("%I64d",&n);
for(i = ; i<=n; i++)
{
scanf("%I64d",&a[i]);
sum+=a[i];
}
for(i = n; i>=; i--)
{
ans*=work(sum-,a[i]-);
ans%=mod;
sum-=a[i];
}
printf("%I64d\n",ans); return ;
}